Prove this polylogarithmic integral has the stated closed form value Question. Prove the following polylogarithmic integral has the stated value:

$$I:=\int_{0}^{1}\frac{\operatorname{Li}_2{(1-x)}\log^2{(1-x)}}{x}\mathrm{d}x=-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$

I was able to arrive at the proposed value for the integral through a combination of educated guessing and after-the-fact numerical checking (the value of the integral was approximately $I\approx0.457621...$), but I'm at a loss for how to derive it rigorously. Thoughts?
 A: Since no answers have been posted, I'll expand on my comment above.
There is a general formula that states $$\sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(x) \log^{r-1}(x) }{1-x}dx   $$ where $$ H_{n}^{(r)} = \sum_{k=1}^{n} \frac{1}{k^{r}} .$$
A proof can be found here.
Making the substitution $u = 1-x$, $$ \sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(1-u) \log^{r-1}(1-u)}{u}dx .$$
Therefore,  $$ \int_{0}^{1} \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx = 2 \zeta (3) \zeta (2)  - 2 \sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} .$$
To simplify the evaluation of that Euler sum slightly, I'm first going to evaluate $ \displaystyle \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}}$ and then use the identity $$\sum_{n=1}^{\infty} \frac{H_{n}^{(r)}}{n^{q}} + \sum_{n=1}^{\infty} \frac{H_{n}^{(q)}}{n^{r}} = \zeta(r) \zeta(q) + \zeta(r+q) . \tag{1} $$ 
Consider $$f(z) = \frac{ \pi \cot (\pi z) \ \psi_{1}(-z)}{z^{3}} $$ where $\psi_{1}(z)$ is the trigamma function.
The function $f(z)$ has poles of order $3$ at the positive integers, a pole of order 6 at the origin, and simple poles at the negative integers.
On the sides of a square with vertices at $\pm \left( N+ \frac{1}{2} \right) \pm i \left( N+ \frac{1}{2} \right)$, $\cot (\pi z)$ is uniformly bounded.
And when $z$ is large in magnitude and not on the positive real axis, $\psi_{1}(-z)$ is approximately $ \displaystyle - \frac{1}{z}$.
So as $N \to \infty$ through the integers, $ \displaystyle \int f(z) \ dz$ will vanish on all four sides of the square.
Therefore, $$ \sum_{n=-\infty}^{\infty} \text{Res} [f(z), n] = 0.$$
The Laurent expansion of $\psi_{1}(-z)$ at the positive integers (including $0$) is $$ \psi_{1}(-z) = \frac{1}{(z-n)^{2}} + \sum_{m=0}^{\infty} (m+1) \left( (-1)^{m} H_{n}^{(m+2)} + \zeta(m+2) \right) (z-n)^{m} . $$
And the Laurent expansion of $\pi \cot \pi z$ at the integers is 
$$ \pi \cot (\pi z) = \frac{1}{z-n} - 2 \sum_{m=1}^{\infty} \zeta(2m) (z-n)^{2m-1} .$$
So at the positive integers, $$f(z) = \frac{1}{z^{3}} \left(\frac{1}{(z-n)^{3}} + \frac{H_{n}^{(2)} - \zeta(2)}{(z-n)} + \mathcal{O}(1) \right) $$ 
which implies
$$ \begin{align} \text{Res} [f(z), n] &= \text{Res} \left[\frac{1}{z^{3}(z-n)^{3}}, n \right] + \text{Res} \left[\frac{H_{n}^{(2)}-\zeta(2)}{z^{3}(z-n)}, n \right] \\ &= \frac{6}{n^{3}} + \frac{H_{n}^{(2)}}{n^{3}} -\frac{\zeta(2)}{n^{3}} . \end{align} $$
At the negative integers,
$$\text{Res} [f(z), -n] = - \frac{\psi_{1}(n)}{n^{3}} = \frac{H_{n-1}^{(2)} - \zeta(2)}{n^{3}} = \frac{H_{n}^{(2)}}{n^{3}} - \frac{1}{n^{5}} - \frac{\zeta(2)}{n^{3}} .$$
And at the origin,
$$ f(z) = \frac{1}{z^{6}} - \frac{\zeta(2)}{z^{4}} + \frac{2 \zeta(3)}{z^{3}} + \left(\zeta(4) - 2 \zeta^{2}(2) \right) \frac{1}{z^{2}} + \Big(4 \zeta(5) - 4 \zeta(3) \zeta(2) \Big) \frac{1}{z} + \mathcal{O}(1)$$
while implies
$$ \text{Res}[f(z),0] = 4 \zeta(5) - 4 \zeta(3) \zeta(2) .$$
Summing up all the residues,
$$6 \zeta(5) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}}  - \zeta(3) \zeta(2) + \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} - \zeta(5) - \zeta(3) \zeta(2) + 4 \zeta(5) - 4 \zeta(3) \zeta(2) = 0$$
which implies
$$ \sum_{n=1}^{\infty} \frac{H_{n}^{(2)}}{n^{3}} = 3 \zeta(3) \zeta(2) - \frac{9}{2} \zeta(5) .$$
Then using $(1)$,
$$\sum_{n=1}^{\infty} \frac{H_{n}^{(3)}}{n^{2}} = \zeta(3) \zeta(2) + \zeta(5) - 3 \zeta(3) \zeta(2) + \frac{9}{2} \zeta(5) = - 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5).  $$
So finally we have
$$ \begin{align} \int_{0}^{1}  \frac{\text{Li}_{2}(1-x)\log^{2}(1-x)}{x} \ dx &= 2 \zeta (3) \zeta (2)  - 2 \Big(- 2 \zeta(3) \zeta(2) + \frac{11}{2} \zeta(5) \Big)  \\ &= 6 \zeta(3) \zeta(2) - 11 \zeta(5) . \end{align}$$
